Showing that R and R^2 are not homeomorphic

So, to prove this, one needs to conclude that there is no homeomorphism between R and R^2. A homeomorphism is a continuous bijection f with a continuous inverse. (Does there exist a bijection at all between these two sets? I assume yes, since they have the same cardinality, but I don't see how to construct it.)

You can kind of envision a direct bijection between (0,1) and (0,1)x(0,1) by given a pair of real numbers, construct a single one by alternating their digits. So the pair (.1275, .9999) becomes .19297959.

There are some issues with well-definedness at points with more than one decimal expansion (for example, .1 vs .099999).

As far as proving no homeomorphism exists, the classic method involves removing the origin from both and seeing what happens to one and not the other

As far as proving no homeomorphism exists, the classic method involves removing the origin from both and seeing what happens to one and not the other

Could you please be a bit more specific, I'm not sure I understand what you meant.

I assume I have to find some topological property which R and R^2 don't share, since if there would exist a homeomorphism between them, this property would be preserved. But I simply can't see it - both spaces are connected and separable, right? I should find some other property.

Well, if R is homeomorphic to R^2, we know that R^2 is connected, too, since continuous functions (and homeomorphisms in particulas) preserve that property. If we remove some x from R now, R\{x} isn't connected anymore. I'm not sure what's the picture. Is we look at the restriction of the mapping to R\{x}, it stays continuous, right? But I only know that it maps connected sets to connected sets. Can it happen for the domain not to be connected, and the image of the map connected?

Well, if R is homeomorphic to R^2, we know that R^2 is connected, too, since continuous functions (and homeomorphisms in particulas) preserve that property. If we remove some x from R now, R\{x} isn't connected anymore. I'm not sure what's the picture. Is we look at the restriction of the mapping to R\{x}, it stays continuous, right? But I only know that it maps connected sets to connected sets. Can it happen for the domain not to be connected, and the image of the map connected?

R\{x} is not connected, and our homeomorphism is still a homeomorphism. The space R\{x} is allegedly homeomorphic to is going to be R2\{f(x)}

In general a domain can be disconnected and have a connected image; for example any constant map. But this can't happen for homeomorphisms (we would hope not, otherwise homeomorphic would be a terribly weak property) and is something that can be proven fairly simply

R\{x} is not connected, and our homeomorphism is still a homeomorphism. The space R\{x} is allegedly homeomorphic to is going to be R2\{f(x)}

In general a domain can be disconnected and have a connected image; for example any constant map. But this can't happen for homeomorphisms (we would hope not, otherwise homeomorphic would be a terribly weak property) and is something that can be proven fairly simply

OK, we proved that R\{x} and R\{f(x)}^2 are not homeomorphic, but how does this imply that R and R^2 are not homeomorphic? Obviuosly both R\{x} and R\{f(x)}^2 are not connected, and our homeomorphism stays a homeomorphism from a disconnected set to a disconnected set. Should we prove that a homeomorphism doesn't in general map a disconnected set into a disconnected set? Or is it simply a game of logic - if we have the implication (which is true) "A is connected => f(A) is connected under a homeomorphism f", then its negation is false, i.e. "A is not connected => f(A) is not connected under a homeomorphism f"? (btw, I'm not really sure my logic is right here) Thanks for your patience :)

Not R\{f(x)}2 (which doesn't even make sense), we're looking at R2\{f(x)}; the plane minus a point

Suppose f is a homeomorphism from R to R2. Then here are the key points:
1) f is also a homeomorphism from R\{x} to R2\{f(x)}
2) R\{x} is disconnected, R2\{x} is connected
3) A connected set cannot be homeomorphic to a disconnected set

Not R\{f(x)}2 (which doesn't even make sense), we're looking at R2\{f(x)}; the plane minus a point

Suppose f is a homeomorphism from R to R2. Then here are the key points:
1) f is also a homeomorphism from R\{x} to R2\{f(x)}
2) R\{x} is disconnected, R2\{x} is connected
3) A connected set cannot be homeomorphic to a disconnected set

You'll have to fill in the details for the proof

You meant R^2\{f(x)} is connected, right? (in 2) )

So, if f : R\{x} --> R^2\{f(x)} is a homeomorphism, then it has a continuous inverse f^-1, and since continuous function preserve connectedness, it follows that R\{x} is connected, which is a contradiction.